19 homework YOO, HEE Due: Mar 4 2008, 4:00 am Question 1, chap 8, sect 5. part 1 of 2 10 points A 1880 kg car starts from rest and accelerates uniformly to 18.7 m/s in 14.1 s . Assume that air resistance remains constant at 321 N during this time. Find the average power developed by the engine. Correct answer: 35.2735 hp (tolerance 1 %). Explanation: P = Fengine vf m = 1880 kg , vi = 0 m/s , vf = 18.7 m/s , t = 14.1 s . The acceleration of the car is a= since vi = 0, so a= 18.7 m/s 14.1 s = 1.32624 m/s2 . vf vf - vi = t t = (321 N)(18.7 m/s) and = 70.547 hp . 1 hp 764 W Question 2, chap 8, sect 5. part 2 of 2 10 points
1
Find the instantaneous power output of the engine at t = 14.1 s just before the car stops accelerating. Correct answer: 70.547 hp (tolerance 1 %). Explanation: The instantaneous velocity is 18.7 m/s and the instantaneous power output of the engine is
Question 3, chap 9, sect 99. part 1 of 2 10 points The universal gravitationan constant is 6.6726 10-11 N m2 /kg2 . Three 2 kg masses are located at points in the xy plane as shown in the figure.
Thus the constant forward force due to the engine is found from F = Fengine - Fair = m a = 321 N + (1880 kg) 1.32624 m/s2 = 2814.33 N . The average velocity of the car during this interval is vf + vi , vav = 2 so the average power output is P = Fengine vav = Fengine vf 2 18.7 m/s = (2814.33 N) 2
52 cm
Fengine = Fair + m a
35 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 2.39197 10-9 N (tolerance 1 %). Explanation:
1 hp 764 W
= 35.2735 hp .
Let : m = 2 kg x = 35 cm , y = 52 cm .
and
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am Basic Concepts: Newton's Law of Gravitation: m1 m2 Fg = G r2 We calculate the forces one by one, and then add them using the superposition principle. The force from the mass on the right is pointing in the x direction, and has magnitude mm f1 = G 2 x G m2 = x2 (6.6726 10-11 N m2 /kg2 )(2 kg)2 = (0.35 m)2 = 2.17881 10-9 N .
2
Note: As you see, this force is very small. If the masses in the picture are standing on a table (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. Question 4, chap 9, sect 99. part 2 of 2 10 points At what angle (-180 < < 180 ) from the horizontal (x) axis will the resultant force point? Correct answer: 24.3721 (tolerance 1 %). Explanation: The angle shown in the figure in Part 1 is = arctan f2 f1 (9.87071 10-10 N) = arctan (2.17881 10-9 N)
The other force is pointing in the y direction and has magnitude mm f2 = G 2 y G m2 = y2 (6.6726 10-11 N m2 /kg2 )(2 kg)2 = (0.52 m)2 = 9.87071 10-10 N .
= 24.3721 .
Question 5, chap 9, sect 4. part 1 of 2 10 points Given: G = 6.672 10-11 N m2/kg2 . The planet Mars requires 2.35 years to orbit the sun, which has a mass of 2 1030 kg, in an almost circular trajectory. Calculate the radius of the orbit of Mars as it circles the sun. Correct answer: 2.64784 1011 m (tolerance 1 %). Explanation: Fr = G and = 2 T Mm = m ar = m 2 r , r2
f2
F f1
Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythagoras' theorem as well: F =
2 2 f1 + f2
= (2.17881 10-9 N)2 + (9.87071 10-10 N)2 = 2.39197 10-9 N .
1 2
=
2 (7.41096 107 s) = 8.47823 10-8 rad/s .
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am Thus, GM 2
1 3
3
r=
2 1030 kg = (8.47823 10-8 rad/s)2
1 3
(6.672 10-11 N m2 /kg2 ) 3 = 2.64784 1011 m . Question 6, chap 9, sect 4. part 2 of 2 10 points
1
Calculate the orbital speed of Mars as it circles the sun. Correct answer: 22449 m/s (tolerance 1 %). Explanation: v = r = (8.47823 10-8 rad/s) (2.64784 1011 m) = 22449 m/s .
Question 7, chap 9, sect 99. part 1 of 2 10 points Two satellites A and B orbit the Earth in the same plane. Their masses and radii have the relationships mB = 6 mA and rB = 3 rA . B
vB = 3 vA vB 1 4. = vA 3 vB = 2 5. vA vB 6. =2 vA vB 1 7. = correct vA 3 1 vB = 8. vA 2 vB 9. =9 vA vB 1 10. = vA 2 Explanation: Basic Concepts: Force of gravity between two masses m1 and m2 at a distance r m1 m2 , Fg = G r2 where G is the gravitational constant. Circular motion v2 ar = , r for the radial (centripetal) acceleration. Note: v is the tangential speed. Solution: Since the force of gravity is responsible for holding a satellite in its orbit, the orbital centripetal force is equal to the force of gravity 3. v2 Mm =G 2 , r r where again M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (from the Earth's center). Thus the tangential speed v of an orbit at distance r is, Fr = m GM . r Note: The speed is independent of m. Thus the ratio v= vB = vA GM rB GM rA
rB r A
A
What is the ratio of the orbital speeds 1. vB =3 vA vB 1 2. = vA 9
vB ? vA
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am rA . rB And since rB = 3 rA, we have = vB = vA rA 1 = . 3 rA 3 Question 9, chap 9, sect 99. part 1 of 1 10 points
4
Question 8, chap 9, sect 99. part 2 of 2 10 points Let the distance of the satellite A from the center of the Earth be rA = 10 R, where R is the radius of the Earth. Denote the gravitational acceleration at the surface of the Earth by g. The gravitational acceleration due to the Earth at satellite A is given by 1. gA = 2. gA = 3. gA = 4. gA = 5. gA = 6. gA = 7. gA = 8. gA = g 9 g 3 g 10 g 121 g 81 g 11 g 10 g 11
The planet Mars has a mass 6.9 1023 kg and radius 3.5 106 m. The gravitational constant is 6.67259 10-11 N m2 /kg2 . What is the acceleration of an object in free-fall near the surface of Mars? Correct answer: 3.75844 m/s2 (tolerance 1 %). Explanation: Let : M = 6.9 1023 kg , G = 6.67259 10 R = 3.5 106 ,
and m N m2 /kg2 .
-11
Near the surface of Mars, the gravitation force on an object of mass m is F =G Mm , R2
where M is the mass of Mars and R is the radius of Mars. Therefore the acceleration of a free-fall object is a= F m
=G
9. gA = g g correct 100 Explanation: Since GM , g= R2 then at rA , 10. gA = gA = GM 2 rA GM = (10 R)2 g . = 100
M R2 = (6.67259 10-11 N m2 /kg2 ) 6.9 1023 kg (3.5 106 m)2 = 3.75844 m/s2 .
Question 10, chap 9, sect 99. part 1 of 1 10 points An object has a weight W when it is on the surface of a planet of radius R. What will be the gravitational force on the object after it has been moved to a distance of 4 R from the center of the planet? 1. F = 4 W
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am 1 2. F = W 4 1 3. F = W correct 16 4. F = W Explanation: On the surface of the planet, GM m W= R2 G M m = W R2 . When the object is moved to a distance 4R from the center of the planet, the gravitational force on it will be GM m F = (4 R)2 GM m = 16 R2 1 GM m = 16 R2 1 W. = 16 Question 11, chap 9, sect 99. part 1 of 1 10 points A cylindrical habitat in space 4.7 km in diameter and 26.2 km long has been proposed (by G.K. O'Neill, 1974). Such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. This would all be held in place by rotation of the cylinder about its long axis. The acceleration of gravity is 9.8 m/s2 . How fast would the cylinder have to rotate to imitate the Earth's gravitational field at the walls of the cylinder? Correct answer: 0.0645772 rad/s (tolerance 1 %). Explanation: The radius of a cylinder with d = 4.7 km is d r= 2 4.7 km 1000 m 1km = 2 = 2350 m . 5. F = 16 W
5
To simulate the Earth's gravitational field g = 9.8 m/s2 at the walls of the cylinder the following relation must hold: g = 2 r = v2 . r
Therefore the angular speed of the cylinder has to be = = g r
9.8 m/s2 2350 m = 0.0645772 rad/s .
Question 12, chap 9, sect 99. part 1 of 3 0 points Given: A solar system, different from but similar to our Sun, Earth, and Moon, where Mearth = 4.04 1024 kg Rearth = 5.77 106 m Msun = 2.12 1030 kg sun Rearth = 1.43 1011 m Mmoon = 8 1022 kg moon Rearth = 4 108 m G = 6.67259 10-11 N m2 /kg2 .
Sun
Moon
Planet
By what percentage does the weight of the 71.2 kg woman standing in a total eclipse of the Sun decrease due to the Sun's gravitational force (neglecting the Moon's gravitational force) when compared to her weight due to the Earth's mass alone? Correct answer: 0.0854415 % (tolerance 1 %). Explanation:
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am Her weight on Earth (with the accepted value for the Earth's mass and radius) is normally W = m g = 71.2 kg 9.8 m/s2 = 697.76 N , but since this solar system is different from ours, we cannot use our value g = 9.8 m/s2 . The Earth's gravitational force is Fearth G m Mearth = 2 Rearth = (6.67259 10-11 N m2 /kg2 ) (71.2 kg) (4.04 1024 kg) (5.77 106 m)2 = 576.506 N , The Sun's gravitational force is Fsun = G m Msun 2 rS-E where
moon rM -E = Rearth - Rearth = (4 108 m) - (5.77 106 m) = 3.9423 108 m .
6
By what percentage does the weight of the 71.2 kg woman standing in a total eclipse of the Sun decrease due to the Moon's gravitational force (neglecting the Sun's gravitational force) when compared to her weight due to the Earth's mass alone? Correct answer: 0.00042419 % (tolerance 1 %). Explanation: The Moon's gravitational force is Fmoon = G m Mmoon 2 rM -E
= (6.67259 10-11 N m2 /kg2 ) (71.2 kg) (8 1022 kg) (3.9423 108 m)2 = 0.00244549 N ,
= (6.67259 10-11 N m2 /kg2 ) (71.2 kg) (2.12 1030 kg) (1.42994 1011 m)2 = 0.492576 N , where
sun rS-E = Rearth - Rearth = (1.43 1011 m) - (5.77 106 m) = 1.42994 1011 m ,
The percentage decrease is 0.00244549 N Fmoon 100 = Fearth 576.506 N = 0.00042419% . Alternative Solution: The percentage decrease is Mmoon Rearth Fmoon = Fearth Mearth rM -E = (0.019802) (0.0146361)2 100 = 0.00042419% .
2
The percentage decrease is Fsun 0.492576 N 100 = Fearth 576.506 N = 0.0854415% . Alternate Solution: crease is The percentage de2
Msun Fsun Rearth = Fearth Mearth rS-E = (524752) (4.03513 10-5)2 100 = 0.0854415 % . Question 13, chap 9, sect 99. part 2 of 3 0 points
Question 14, chap 9, sect 99. part 3 of 3 0 points Notice: The gravitational force due to the Sun is much larger than the gravitational force due to the Moon. In your consideration, which statement is correct?
homework 19 YOO, HEE Due: Mar 4 2008, 4:00 am 1. The Sun is primarily responsible for the tides. 2. The Moon is primarily responsible for the tides. correct Explanation: Hint: Calculate the fractional change in the gravitational force due to the rotation of the Earth. The magnitude of the gravitational force is mM F =G 2 , r and F dF r dr F -2 G Using Fmoon = G m Mmoon , (Lmoon )2 earth m Mmoon r , (Lmoon )3 earth
7
Using the same technique, we have for the Sun F =4 F =4 Msun Mearth Rearth Lsun earth
3
6.37 106 m 1.496 1011 m = 1.02814 10-7 ,
1.991 1030 kg 5.98 1024 kg
3
where Lsun = 1.496 1011 m is the distance earth from the Earth to the Sun. Obviously, the Moon's gravitational force is primarily responsible for the tides, since 2.2473 10-7 > 1.02814 10-7 .
mM r . r3
However, the angular acceleration due to the moon's orbital motion about the center of mass of the Earth-Moon system is the largest contributing factor which causes the tides.
Fmoon = -2 G Fearth = G
m Mearth , 2 Rearth
since r = 2 Rearth (the Earth's diameter), we have for the Moon Fmoon = Fearth -2 G m Mmoon (2 Rearth) (Lmoon )3 earth m Mearth G 2 Rearth Rearth Lmoon earth
3
F =4 F =4
Mmoon Mearth
6.37 106 m 3.84 108 m = 2.2473 10-7 ,
7.36 1022 kg 5.98 1024 kg
3
where Lmoon = 3.84 108 m is the distance earth from the Earth to the Moon.